How do you use substitution to solve the pair of equations: #y=2x-3# and #x^2+y^2=16#?

1 Answer
Alan P.
Dec 6, 2016

#(x,y)~~color(green)(""(2.885,2770)) or color(green)(""(-0.485,-3.970))#

Explanation:

Given
[1]#color(white)("XXX")y=2x-3#
[2]#color(white)("XXX")x^2+y^2=16#
Using [1] we know that we can substitute #(2x-3)# for #y# in [2]
[3]#color(white)("XXX")x^2+(2x-3)^2=16#

[4]#color(white)("XXX")rarr x^2+4x^2-12x+9=16#

[5]#color(white)("XXX")rarr5x^2-12x-7=0#

Using the quadratic formula:
[6]#color(white)("XXX")x=(12+-sqrt((-12)^2-4 * 5 * (-7)))/(2 * 5)#

simplifying (and using a calculator)
#color(white)("XXX"){:(x~~2.885," or ",x~~-0.485):}#
using [1] #rarr#
#color(white)("XXX"){:(y~~2.770," or ", y~~-3.970):}#

The image of the given relations (below) indicates that these results are reasonable.
enter image source here

Related questions
See all questions in Systems Using Substitution
Impact of this question
936 views around the world
You can reuse this answer
Creative Commons License